What is the integral of+(4y^2-7y-12)/(y(y+2)(y-3)) ?

The integral of+(4y^2-7y-12)/(y(y+2)(y-3)) is 2ln|y|+9/5 ln|y+2|+1/5 ln|y-3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of+(4y^2-7y-12)/(y(y+2)(y-3))

\int + \frac{4 y ^{2} - 7 y - 12}{y ( y + 2 ) ( y - 3 )} d y

2 ln ∣ y ∣ + \frac{9}{5} ln ∣ y + 2 ∣ + \frac{1}{5} ln ∣ y - 3 ∣ + C

Solution steps

\int \frac{4 y ^{2} - 7 y - 12}{y ( y + 2 ) ( y - 3 )} d y

Take the partial fraction of

\frac{4 y ^{2} - 7 y - 12}{y ( y + 2 ) ( y - 3 )} : \frac{2}{y} + \frac{9}{5 ( y + 2 )} + \frac{1}{5 ( y - 3 )}

= \int \frac{2}{y} + \frac{9}{5 ( y + 2 )} + \frac{1}{5 ( y - 3 )} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2}{y} d y + \int \frac{9}{5 ( y + 2 )} d y + \int \frac{1}{5 ( y - 3 )} d y

\int \frac{2}{y} d y = 2 ln ∣ y ∣

\int \frac{9}{5 ( y + 2 )} d y = \frac{9}{5} ln ∣ y + 2 ∣

\int \frac{1}{5 ( y - 3 )} d y = \frac{1}{5} ln ∣ y - 3 ∣

= 2 ln ∣ y ∣ + \frac{9}{5} ln ∣ y + 2 ∣ + \frac{1}{5} ln ∣ y - 3 ∣

Add a constant to the solution

= 2 ln ∣ y ∣ + \frac{9}{5} ln ∣ y + 2 ∣ + \frac{1}{5} ln ∣ y - 3 ∣ + C

\int \frac{1}{x ^{0.8}} d x

\int 4 x^{2} e^{3 x^{2}} d x

\int (\frac{3}{x ^{4}} - \frac{2}{x}) d x

\int sin^{2} (y) cos^{4} (y) d y

\int \frac{1}{2} \cdot e^{- \frac{x}{2}} d x

What is the integral of+(4y^2-7y-12)/(y(y+2)(y-3)) ?
The integral of+(4y^2-7y-12)/(y(y+2)(y-3)) is 2ln|y|+9/5 ln|y+2|+1/5 ln|y-3|+C